Answer:
the answer is 0.
Step-by-step explanation:
The answer is 0, I had a question like this before
There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
700a+1200r=1000000 (then plug in the a)
700(1000) +1200r=1000000 (simplify)
700000+1200r=1000000 (subtract 700000 from both sides)
1200r=300000 (divide by 1200 on both sides)
r=250
250 pounds
Answer:
i think it would be A
Step-by-step explanation:
